exp.#8

we had to set up a circut and check the voltage at vce,vbe,ic,ib useing
1NPN diode bc547
1resistor 470
vary resistors2.2K,47k,220k,270,1M
then i had to record the voltage at the trasistor at vce,vbe,ic,ib for each of the resistors.

the voltage at vce was geting bigger as we put bigger resistance because we need more preusser to push through the bigger resistor.

the voltage at vbe didnt chage much as we didnt chage the resistance.

The voltage at ib keep droping as we put bigger resistors becase ther was less current.
the ic chaged because ther was less current going throgh, and we getting bigger resistance on the base

then we out our reading frm ic and vce on a graph and create a loadline.

circuit number 3

we had to build a ocean sesnior, we got given a plan sheet had to figer out the the all diffrence resistors we new the value of R6 at 10k

I stared from D2 the !st diode after V.S we needed to kno what was the V.D whichh is 0.6 we end up with 11.4v after the diode to R5 we didnt have the value use smple ohms law to figer out the value.We new the Z diode was at 9.1v there is a capacitor ther aswell to keep the voltage stayble. I worked out that R6 V.D is gota be 8.4V because we need 0.63v (I= V/R = 0.00085a). we kno from kurchofs law that in a series circuit the current shold always be the same at any point.
Then again we used ohms law to figer out R8 R= V/I = 740ohms
same for R7 we had the voltage after R8 which was 0.23v (R=V/I).

R5 13411 ohms
R6 10k ohms
R7 270ohms
R8 740ohms

then we go to the paraelle side of the circuit and had to work out R2,R3,R4.
we new 11.4 is come to the output of the op amp going to the L.E.D haveing the voltdrop of 1.8v, useing ohms vs-vd=9.6 and we kno the current is 9.5mA V/I=R 9.6V/9.5mA = 1010ohms. we did same thing for R4.
for R3 we had a diode and a l.e.d 0.6vd for the diode and 1.8vd for l.e.d which gives us 9v R=V/I 9V/9.5MA=947OHMS.
I KNO HOW THIS CIRCUIT WORKS IN THERORY BUT DIDNT BUILD IT

WORKING OUT:
R8 v/i=r 0.63/0.00085 = 740ohms
R7 " " 0.23/0.00085 = 270ohms
R2 " " vs-vd = 9.6,9.6/9.5mA = 1010ohms
R4 " " " " " " " " " "
R3 vd+vd = 2.1 vs-tvd 9 v/i=r 9v/9.5mA=947ohms

exp.5

we had to build a small circuit using these components:

3 diffrent resistors




we 1st calculated how much time it would take the capacitor to charge up. them measure the time take to reach applied voltage with a oscilloscope then make a graft we use 100 uf for each resistance at 2v.

we had to build a small circuit using these components:

3 diffrent resistors x2 1k 0.1k 0.47k








we 1st calculated how much time it would take the capacitor to charge up. them measure the time take to reach applied voltage with a oscilloscope then make a graft we use 100 uf for each resistance at 2v.

graph for exp.5

circuit #	capacitance        (uf)	resistance	calculated time	observed time  (ms)
1	100	1k	0.5	0.6
2	100	0.1k	0.5	49
3	100	0.47k	0.234	236ms
4	330	1k	1.63	175ms

exp no.6

we had to use a multimeter and check a BJT transistor  and lean how to tell between PNP and NPN also be able to fine out which of the three post are the emitter colcetor and  base.
Then we analised that the BJT will not work if you dont get the POLARITIES right, we found that by useing a multimeter.  

NPN TYPE

vbe 0.825
veb 0.829
vbc O/L
vcb 0.822
vce O/L
vec O/L

PNP TYPE

vbe 0.825
veb O/L
vbc 0.822
vcb O/L
vceO/L
vecO/L

ex no4

We had to build a circuit with 1 diode 1 zener diode and one resistors at 1k ohms voltage supply was 10v. We then had to do some voltage drop test then change the voltage to 15v and do the same thing.from the test we found out voltage was almost the same but the current would change. for 5vs at the resistor it was 4.61 and for 10v it was 9.56. then we use ohms law to get the current
V = AxR
4.61/1000=.004

no.2
V= A x R
9.56/1000=.009

when we put the voltage higher the currant gets higher at the resistor but almost the same everywhere else in the circuit

exp no7

i had to bulid a small sircuit with one signal npn type tresistors and 2 resistors with @1kohms and another @ 10k ohms, we then had to test a transistor by doing a voltdrop test between the base and the emmiter. the resultwas 0.79v which ment that is was at the saturated point , i kew this result by looking at the region chart, when the result is over 1v it is in active region and when the result is more then it will not operate                          step 2
then we did a voltdrop test between the colector and the emmiter, the result was 0.5v at the saturated point.its fully open thats why we only need 0.5v to pust it throgh.
at the VCE of 3v the power disspated by the transisots is 45w we figured this out by useing these calcultions p/w = A x V
15 x 3 = 0.45w
 then we had to work out the beta of the transistor useing this formula
B = ic/ib

this is the kind of chart we use to find out what region the transistor is working in

 

      Components use:

• capacitor X 2
• rectifier diode X 2
• l.e.d
• Z diode
• voltage regulator
• 3 X resistor @ 1800ohms 2700ohms 8200ohms

How it works:
12v going through diode then to z.d then  to the cap (store voltage) and then input of the voltage reg with the feed back resistor for protection its regulates the voltage down to 5v. we also have a gound out put, 12v out put, l.e.d to show that it is working. at 1st it didnt work as it was ment to work because D12 was shorting

we worked out the resistor by using ohms law R = V/I

volt drops

• diode13 0.6v
• cap12.2v
• volt reg 5v
• R1 3.9v
• led 1.78
• R3 3.3v
• R2 1.14v

        components use

• transistor X 2  PNP TYPE

•         l.e.d X 2 
•         560 ohms resistors X 2
•         860 ohms resistor X 2.

  We started of making the circuit on a lock master to plan it out 1st. before that we had to work out what resistance to use. after hooking up the components supplied 12v but the clear l.e.d did not work, i looked at the bottom peace their was a lose connection on R3 fix that and the light was glowing like never has.

Formulas used
B = IC/IB
V - VD
V=R x I
VP-VDC/IB=R

how it was working.

the volts goes though the circuit and makes the current flow, the 12v is going through R14 R15 then goes to the l.e.d then to the collector. the 5volts  PWM going to R13 R16 and them going to the base of the transistor which then open the gate to C and E of the transistor to let flow through.

volt drop test.
R14 R15 v.d 9.6v
R13 R16 v.d 3.6v (vs 5v)
L.E.D v.d 1.76
L.E.D (CLEAR) v.d 2.1v
V.B.C 12.3

         

experiment no.3

we had to build a circuit on a breadboard using the following components, 2 100ohms resisters, 1  5V, 1 400mw zen er diode (ZD)

R = 100 OHMS AND RL = 100OHMS VS = 12V
We then had to work out the value of vz which is 5v

when we vary the VS from 10 to 15vs the value of the VZ is almost the same

• At 10vs the voltage drop is 4.2v and at 15vs the VD was 4.6 

• this kind of circuit is used to vloatge regulate voltage, z diode  to keep the voltage stabile.